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Counting permutations uses combinatorics and factorials

Example: For \( n \) items, the number of permutations is equal to \( n! \) (factorial of \( n \))

How to count distinct permutations?

Having a repeated item involves a division of the number of permutations. Count the number of permutations of these repeated items.

Example: DCODE letters have \( 5! = 120 \) permutations but contain the letter D twice (these \( 2 \) letters D have \( 2! \) permutations), so divide the total number of permutations \( 5! \) by \( 2! \): \( 5!/2!=60 \) distinct permutations.

How to remove the limit when computing permutations?

Permutations makes exponential values witch needs huge computing servers with huge memory cells, so the generation must be paid.

Source code

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> [News]: Discover the next version of dCode Permutations!

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News]: Discover the next version of dCode Permutations!